NATIONAL
SENIOR CERTIFICATE
GRADE 12
MATHEMATICAL LITERACY P1
NOVEMBER 2009
MEMORANDUM
MARKS: 150
Symbol
M
MA
CA
A
C
S
RT/RG
SF
O
P
R
Explanation
Method
Method with accuracy
Consistent accuracy
Accuracy
Conversion
Simplification
Reading from a table/Reading from a graph
Correct substitution in a formula
Opinion/Example
Penalty, e.g. for no units, incorrect rounding off etc.
Rounding off
This memorandum consists of 16 pages.
_________________________
EXTERNAL MODERATOR
MR M.A. HENDRICKS
Copyright reserved
________________________________
INTERNAL MODERATOR
MRS J. SCHEIBER
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QUESTION 1 [26]
Ques Solution
1.1.1
DoE/12 November 2009
Explanation
464 : 128
( ÷ 16)
29 : 8 9A
AS
12.1.1
1A solution
(1)
12.1.1
1.1.2
1A solution
1CA rounding off
379
= 1,516 9A
250
≈ 1,52 9CA
(2)
ANSWER ONLY – FULL MARKS
1,5 or 1,51 – 1 mark
Any other incorrect answer - 0
1.1.3
OR
9A
57 21 + 57
7+
=
3
3
9A
7+
1
(57)
3
= 7 + 19
9A
78
3
= 26
9A
=
= 26 9CA
1.1.4
12.1.1
1A square root
1A simplifying brackets and
dividing
1CA simplification
9CA
(3)
ANSWER ONLY – 2 marks
12.3.2
1,25 × 1 000 m l 9M
1M multiplying by 1 000
= 1 250 m l 9A
1A accurate conversion
(2)
ANSWER ONLY – FULL MARKS
No penalty if units are omitted
1.1.5
16
16% of 1 255 kg =
× 1 255 kg
100
9A
= 200,8 kg
New amount
= 1 255 kg + 200,8 kg
9CA
= 1 455,8 kg
9M
12.1.1
12.3.1
1M calculating %
1A solution
1CA increase in %
OR
16% increase = 1,16
1A total %
9A
New amount = 1,16 × 1 255 kg
= 1 455,8 kg
9M
9CA
1M multiplying
1CA solution
(3)
ANSWER ONLY – FULL MARKS
No penalty if units are omitted
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Solution
1.1.6
$1 = R10,52
3
NSC – Final Memorandum
DoE/12 November 2009
Explanation
AS
12.1.3
1M multiplying
9M
$1 215,00 = R10,52 × 1 215,00
1CA simplification
(2)
= R12 781,80 9CA
ANSWER ONLY – FULL MARKS
12.1.1
R 399,00 9MA
30
1.2.1
= R13,30
1MA dividing
1A simplification
9A
OR
Total number of grams in a box = 500 g × 30
= 15 000 g
R399,00
× 500
15 000
9A
= R13,30
9MA
1MA multiplying
Cost of 500 g =
1A simplification
(2)
ANSWER ONLY – FULL MARKS
1.2.2
1 or 100% or certain 99A
12.4.5
2A correct probability
(2)
ANSWER WRITTEN AS A RATIO – 1
mark
1.2.3
9SF
9
Temp in oF = × 225o + 32o
5
= 405 oF + 32 oF
9S
= 437 oF
≈ 435 oF 9CA
1SF substitution in formula
12.3.2
1S simplification
1CA rounded off to 5 degrees
(3)
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Ques Solution
1.3.1
DoE/12 November 2009
Explanation
Cost price of 1 orange =
R 9,00
12
AS
12.1.1
9M
= R0,75 9CA
1M division by 12
OR
1CA simplification
R9,00 × 100
9M
12
9CA
= 75 cents
Cost price of 1 orange =
1.3.2
(2)
ANSWER ONLY – FULL MARKS
0,75 with no units – 2 marks
75 with no units – 1 mark
12.1.3
1 dozen oranges sell for R12,00 9A
1A selling price for 1 dozen
Profit = R12,00 – R9,00
1CA difference
= R3,00 9CA
OR
Selling price per orange = 100 cents
Cost price per orange = 75 cents
Profit per orange = 25 cents
1A profit per orange
9A
Profit per dozen orgaanges = 25 cents × 12
= 300 cents
= R3,00
9CA
1CA profit per dozen
(2)
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Ques Solution
1.3.3
DoE/12 November 2009
Explanation
Cost = 108 × R0,75 9CA
= R81,00
9CA
AS
12.1.1
1CA cost per orange
1CA cost for 108 oranges
OR
12 oranges cost R9,00
108 × R 9,00
12
= R81,00 9CA
108 oranges =
1M finding number of dozens
9M
1CA cost for 108 oranges
OR
Number of dozen =
108
=9
12
9M
1M dividing
Cost = 9 dozen × R9,00 per dozen
= R81,00
9CA
1CA cost for 108 oranges
(2)
ANSWER ONLY – FULL MARKS
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QUESTION 2 [31]
Ques Solution
2.1.1
D = 10 cm
DoE/12 November 2009
Explanation
9A
AS
12.3.1
1A doubling the radius
(1)
2.1.2
L = 29,5 cm – 2,5 cm – 2,5 cm 9M/A
9CA
= 24,5 cm
12.3.1
1MA reducing 29,5 cm
1CA length of certificate
(2)
ANSWER ONLY – FULL MARKS
Only subtract 2,5 once – 1 mark
Use the width – 1 mark
Using a length = 29,5 cm and having
an answer less than 29,5cm - 1 mark
12.3.1
2.1.3
A= πr
2
= 3,14 × (5 cm) 2 9SF
= 78,5 cm 2 9CA
9A
1SF/CA substitution in
formula (CA from 21.1.0
1CA simplifying
1A unit
(3)
ANSWER ONLY – FULL MARKS
Accept
π=
22
7
or
π
on the
calculator
2.1.4
P = 2 (29,5 cm + 21 cm)9SF
= 101 cm 9CA
12.3.1
1SF substitution in formula
1CA simplifying
(2)
ANSWER ONLY – FULL MARKS
2.1.5
A = 29,5 cm × 21 cm
9SF
12.3.1
1SF substitution in formula
1CA simplifying
= 619,5 cm
2
9CA
(2)
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Solution
Explanation
315 : 1 050 9MA
2.2.1
=
3 : 10
DoE/12 November 2009
AS
12.1.1
1MA ratio in correct order
9CA
1CA simplifying
(2)
ANSWER ONLY – FULL MARKS
1 mark if one of the numbers is 1
Accept notation
3
10
but refer to
question 1.1
2.2.2
12.1.1
9A
2
× 315 guests
7
= 90 guests
9CA
1A correct fraction
1CA simplifying
(2)
CORRECT ANSWER ONLY – FULL
MARKS
12.1.1
2.2.3*
1 litre concentrate makes 5 litres of juice 9MA
5 litres concentrate makes 5 × 5 l
= 25 l
9CA
1MA dilution ratio
1CA simplifying
OR
Number of litres of juice = 4 × 5 l + 1 × 5 l
9MA
= 20 l + 5 l
= 25 l
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9CA
1MA dilution ratio
1CA simplifying
(2)
ANSWER ONLY – FULL MARKS
20 l – 1 mark
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Ques
Solution
Explanation
2.3.1
Eastern Cape or A 9A
1A correct province
AS
12.4.4
(1)
12.4.4
2.3.2
D = 100% – 15% – 6% – 13% –50%
= 16 % or 0,16 or
2.3.3
9MA
16 9CA
100
1MA setting up model
1CA simplifying
(2)
ANSWER ONLY – FULL
MARKS
12.4.4
Gauteng or B 9CA
1CA correct province
(1)
Check the answer to D (2.3.2)
2.3.4
9MA 9MA
18
× 88 144 vehicles OR 0,18 × 88 144
100
= 15 865,92 vehicles 9CA
≈ 15 866 vehicles 9R
1MA 18% of vehicles
stolen
1MA correct no. of vehicles
12.1.1
12.4.4
1CA simplifying the
product
1R rounding
(4)
ANSWER ONLY – FULL
MARKS
12.2.3
2.4.1
(a)
R750 9RG
2.4.1
(b)
Loss
2.4.1
(c)
10 99RG
2.4.2
Percentage profit =
1 RG reading from graph
(1)
12.2.3
9A
1A
(1)
12.2.3
2RG reading from graph
(2)
12.1.3
12.2.1
Profit
× 100%
Expenses
9SF
R 400
=
× 100%
R 850
1SF substitution into
formula
= 47,0588 ...% 9S
1S simplification
≈ 47,1% 9R
1R rounding off
(3)
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QUESTION 3 [19]
Ques Solution
3.1.1
Explanation
17 years 9A
AS
12.4.3
1A modal age
(1)
3.1.2
12.4.3
17 years 9A
1A median
(1)
12.4.3
3.1.3
Mean
9M
1M sum of values
=
16 + 16 + 16 + 17 + 17 + 17 + 17 + 17 + 18 + 18 + 19 + 19 + 19 + 20 + 22
15 9MA
=
268
years
15
= 17,8666 … years 9CA
= 17,87 years
1MA dividing by
size of sample
1CA simplifying
9R
1R rounding off
(4)
ANSWER ONLY –
FULL MARKS
12.4.3
3.2.1
(a)
3.2.1
(b)
3.2.2
20% 9A
1A lowest
(1)
12.4.3
100% 9A
1A highest
(1)
12.4.2
PERFORMANCE
LEVEL
PERCENTAGE
RANGE
1
2
3
4
5
6
7
0 to 29
30 to 39
40 to 49
50 to 59
60 to 69
70 to 79
80 to 100
TALLY
////
////
////
////
////
////
////
//// /
///
///
//// /
FREQUENCY
4
5
11
8
5
8
11
9A
9A
9A
9A
9A
9A
1A learners in level 1
1A learners in level 2
1A learners in level 3
1A learners in level 4
1A learners in level 5
1A learners in level 6
1A learners in level 7
9A
(7)
If cumulative frequency
is correct, maximum of 3
marks.
If addition shown in
cumulative frequency,
maximum of 4
Ignore cumulative
frequency if both are
given
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Ques
Solution
Explanation
3.3.1
52 learners × 1,6 m2/learner 9M/A
1M/A multiplication
= 83,2 m2
1A simplifying
9A
AS
12.3.1
(2)
ANSWER ONLY – FULL MARKS
3.3.2
96
Number of learners =
1,6
9M
= 60 learners 9A
12.3.1
1M division/correct values
1A solution
(2)
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QUESTION 4 [23]
Ques Solution
DoE/12 November 2009
Explanation
AS
12.2.3
4.1.1
90 km
9C 9R
1C conversion to time 08:30
1R reading from table
(2)
4.1.2
08:45
99RT
2RT reading from table
12.2.3
(2)
4.1.3
(a)
12.2.1
120 km 9SF
Speed =
2h
= 60 km/h 9CA
1SF substitution into formula
1CA solution
(2)
ANSWER ONLY – FULL MARKS
4.1.3
(b)
72 minutes = 1,2 hours
12.2.1
9A
1A conversion to hours
1SF substitution into formula
Distance
9SF
= 80 km/h
1,2 h
Distance = 80 × 1,2 km
= 96 km
9CA
1CA solution
OR
OR
60 minutes Æ 80 km
12 minutes Æ
1A distance for 1 hour
9A
12
× 80 km = 16 km
60
72 minutes = 80 km + 16 km = 96 km
9A
9CA
1A distance for 12 minutes
1CA distance for 72 minutes
(3)
ANSWER ONLY – FULL MARKS
80 X 72 - 1 Mark
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Solution
DoE/12 November 2009
Explanation
AS
12.2.2
DISTANCE TRAVELLED AGAINST TIME TAKEN
4.1.4
Mr Lebelo Mr Goldman
120
110
9CA
100
90
Distance in km
80
70
60
1A starting point for Mr Lebelo
at 07:15
99A
50
40
2A plot any 4 points
30
20
1CA joining all points plotted
10
0
07:00
9A
07:15
07:30
(4)
07:45
08:00
08:15
08:30
08:45
09:00
CORRECT GRAPH WITHOUT
PLOTTING INDIVIDUAL
POINTS – FULL MARKS
Time
4.1.5
(a)
1 hour (in terms of Mr Goldman)
9RG
12.2.3
1RG Reading from the graph or
table
¾ hour or 45 minutes (in terms of Mr Lebelo)
(1)
4.1.5
(b)
12.2.3
60 km 9RG9RG
2RG Reading from the graph or
table
(2)
12.2.3
4.1.5
(c)
9RG
9RG
100 km – 90 km
= 10 km
9M
9CA
1M subtraction
1RG reading from graph or table
1RG reading from graph or table
1CA simplifying
(4)
ANSWER ONLY – FULL MARKS
4.2*
9M
9A
Cost of petrol = 10 journeys × 8 l × R8,23 per l
= R658,40 9CA
12.1.3
1A Number of journeys
1M multiplication
1CA simplifying
(3)
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Question 5 [18]
Ques Solution
5.1.1
9A9A
7,51 ; 7,51 ; 7,64 ; 7,71 ; 7,81 ; 7,91 ; 8,05 ; 8,22
DoE/12 November 2009
Explanation
2A ascending order
AS
12.
(2) 4.2
Descending order – 1 mark
Leave off 1 value – 1 mark
5.1.2
9A
7,51 metres
12.
4.3
1A mode
(1)
5.1.3
9M
Range = 8,02 m – 7,23 m
9CA
= 0,79 m
12.
4.3
1M largest – smallest
1CA solution
(2)
ANSWER ONLY – FULL
MARKS
5.1.4
9A
Shortest jump = 7,23 m
7,23 m = 7,23 × 100 cm
9CA
= 723 cm
12.
3.2
1A shortest jump
9C
1C conversion
1CA answer in cm
(3)
ANSWER ONLY – FULL
MARKS
5.1.5
5.1.6
7,64 + 7,82
Median =
m
2
= 7,73 m
9A
Charles
1M method
1A solution
9M
12.
4.3
(2)
ANSWER ONLY – FULL
MARKS
99A
12.
4.1
2A solution
(2)
5.2
V = 9 m × 2,75 m × 0,07 m
= 1,7325 m
≈ 1,733m
3
1SF substitution
9SF
1CA simplification
9CA
3
12.
3.1
1CA rounding off and
correct unit
9CA
(3)
ANSWER ONLY – FULL
MARKS
5.3
August 1991 – October 1968
= 22 years 10 months
≈ 23 years
9CA
9CA
9MA
Also accept
1MA method
1991 – 1968
1CA solution
= 23 years
1CA rounding off
12.
1.1
12.
4.4
(3)
ANSWER ONLY – FULL
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QUESTION 6 [18]
Ques Solution
DoE/12 November 2009
Explanation
6.1.1
18,2% 9RT
6.1.2
9RT
9RT
Difference = 7 908 138 – 5 662 911
AS
12.4.4
1RT reading from table
(1)
= 2 245 227 9CA
2RT reading from table
12.1.1
12.4.4
1CA difference
(3)
ANSWER ONLY – FULL MARKS
6.1.3
(a)
A = 100% – 22,3% – 60,2% – 3,6% 9MA
= 13,9%
9CA
12.1.1
12.4.4
1MA correct values
1CA value of A
OR
A=
1 307 549
× 100% 9MA
9 406 829
= 13,9 % 9CA
6.1.3
(b)
9RT
B = 2 194 066 + 7 908 138 + 1 420 335 + 517 580
= 12 036 739
9CA
(2)
ANSWER ONLY – FULL MARKS
1RT reading from table
12.1.1
12.4.4
1CA finding the sum
(2)
ANSWER ONLY – FULL MARKS
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Ques Solution
Explanation
GRANT TYPES AS A PERCENTAGE OF TOTAL GRANTS
RECEIVED
6.1.4
70
AS
12.4.2
1A Old-age
2007 (accept
18%)
9A
1A Child
support in
2007 (accept
66%)
65
60
55
50
Percentage
45
40
2005
35
1A Disability
in 2007
(accept 12%)
2007
1A Other
2007 (accept
4%)
30
25
9A
20
9A
15
10
9A
5
Other
Disability
Child support
Old-age
0
(take out bold
lines in
English
memo)
Types of grants
(4)
6.2.1
12.1.1
30 9RT
960
9M
=
1
32
OR 0,03 OR 3,13%
9S
1RT reading
correct value
for burial
policy
1M dividing
value by total
1S simplifying
ANSWER ONLY
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(3)
6.2.2
9M
9RT
R960 – R15,45 – R24,50 – R60,00 – R30,00 – R40,00 – R86,40
= R703,65
9CA
12.1.1
1RT correct
values
1M method
1CA
simplifying
(3)
ANSWER ONLY
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QUESTION 7 [15]
Ques
Solution
7.1.1 (a)
DoE/November 2009
Explanation
A = 2 × 3 + 1 9SF
1 SF substitution into formula
1CA value of A
(2)
9CA
=7
AS
12.2.1
ANSWER ONLY – FULL MARKS
7.1.1 (b)
10 = B × 3 + 1
12.2.1
9SF
1 SF substitution into formula
1S simplifying equation
3B = 9 9S
1CA value of B
B = 3 9CA
OR
10 = 3 × 3 + 1
1SF substitution into formula
1S simplifying equation
1CA value of B
(3)
9SF 9S
∴ B = 3 9CA
ANSWER ONLY – FULL MARKS
12.2.3
7.1.2
St Patrick’s College
9RT9RT
2RT reading from the table
(2)
12.3.4
7.2.1
C2 (or 2C)
9RG
1RG reading from the map
(1)
9A
7.2.2** From Kokstad College turn right /NE into
Elliot Street.
Continue
9A
At Barclay Road turn right/ SE.
Kokstad Rugby Club will be on the left.
12.3.4
2A correct directions (Street
name and direction)
(2)
12.3.4
7.2.3** South-east/North-west
9A
1A correct direction
(1)
12.3.3
7.2.4
1 cm represents 20 000 cm
9A
1A scale interpretation
9M1M multiplication
Therefore, 5 cm would represent 5 × 20 000 cm
9S
1S simplification
= 100 000 cm
= 1 000 m
9C 1C conversion
(4)
ANSWER ONLY – FULL MARKS
TOTAL: 150 marks
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